Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in
[link] .

Mean grades for four sororities

Sorority 1

Sorority 2

Sorority 3

Sorority 4

2.17

2.63

2.63

3.79

1.85

1.77

3.78

3.45

2.83

3.25

4.00

3.08

1.69

1.86

2.55

2.26

3.33

2.21

2.45

3.18

Using a significance level of 1%, is there a difference in mean grades among the
sororities?

Let
μ
1 ,
μ
2 ,
μ
3 ,
μ
4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.

Note

This is an example of a
balanced design , because each factor (i.e., sorority) has the same number of observations.

H
0 :
μ1=μ2=μ3=μ4

H
a : Not all of the means
μ1,μ2,μ3,μ4 are equal.

Distribution for the test:F3,16

where
k = 4 groups and
n = 20 samples in total

df (
num )=
k – 1 = 4 – 1 = 3

df (
denom ) =
n –
k = 20 – 4 = 16

Calculate the test statistic:F = 2.23

Graph:

Probability statement:p -value =
P (
F >2.23) = 0.1241

Compare
α and the
p -value:α = 0.01
p -value = 0.1241
α <
p -value

Make a decision: Since
α <
p -value, you cannot reject
H
0 .

Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.

Try it

Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in
[link] .

Gpas for four sports teams

Basketball

Baseball

Hockey

Lacrosse

3.6

2.1

4.0

2.0

2.9

2.6

2.0

3.6

2.5

3.9

2.6

3.9

3.3

3.1

3.2

2.7

3.8

3.4

3.2

2.5

Use a significance level of 5%, and determine if there is a difference in GPA among the teams.

With a
p -value of 0.9271, we decline to reject the null hypothesis. There is not sufficient evidence to conclude that there is a difference among the GPAs for the sports teams.

A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in
[link] .

Tommy's Plants

Tara's Plants

Nick's Plants

24

25

23

21

31

27

23

23

22

30

20

30

23

28

20

Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance.

This time, we will perform the calculations that lead to the
F' statistic. Notice that each group has the same number of plants, so we will use the formula
F' =
n⋅sx–2s2pooled .

First, calculate the sample mean and sample variance of each group.

Tommy's Plants

Tara's Plants

Nick's Plants

Sample Mean

24.2

25.4

24.4

Sample Variance

11.7

18.3

16.3

Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4).
Variance of the group means = 0.413 =
sx–2

Then
MS
between =
nsx–2 = (5)(0.413) where
n = 5 is the sample size (number of plants each child grew).

Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3).
Mean of the sample variances = 15.433 =
s2pooled

Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.

Try it

Notation

The notation for the
F distribution is
F ~
Fdf (
num ),
df (
denom )

where
df (
num ) =
dfbetween and
df (
denom ) =
dfwithin

The mean for the
F distribution is
μ=df(num)df(denom)–1

References

Tomato Data, Marist College School of Science (unpublished student research)

Chapter review

Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the
F statistic from an
F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table.

Formula review

SSbetween=∑​[(sj)2nj]−(∑​sj)2n

SStotal=∑​x2−(∑​x)2n

SSwithin=SStotal−SSbetween

dfbetween =
df (
num ) =
k – 1

dfwithin =
df(denom) =
n –
k

MSbetween =
SSbetweendfbetween

MSwithin =
SSwithindfwithin

F =
MSbetweenMSwithin

k = the number of groups

n
j = the size of the
j
th group

s
j = the sum of the values in the
j
th group

n = the total number of all values (observations) combined

x = one value (one observation) from the data

sx–2 = the variance of the sample means

s2pooled = the mean of the sample variances (pooled variance)

Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The one-way ANOVA results are shown in
[link] .

Group 1

Group 2

Group 3

216

202

170

198

213

165

240

284

182

187

228

197

176

210

201

What is the Sum of Squares Factor?

4,939.2

What is the Sum of Squares Error?

What is the
df for the numerator?

2

What is the
df for the denominator?

What is the Mean Square Factor?

2,469.6

What is the Mean Square Error?

What is the
F statistic?

3.7416

Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in
[link] .

Team 1

Team 2

Team 3

Team 4

1

2

0

3

2

3

1

4

0

2

1

4

3

4

0

3

2

4

0

2

What is
SS
between ?

What is the
df for the numerator?

3

What is
MS
between ?

What is
SS
within ?

13.2

What is the
df for the denominator?

What is
MS
within ?

0.825

What is the
F statistic?

Judging by the
F statistic, do you think it is likely or unlikely that you will reject the null hypothesis?

Because a one-way ANOVA test is always right-tailed, a high
F statistic corresponds to a low
p -value, so it is likely that we cannot accept the null hypothesis.

Questions & Answers

do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?

fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.

Tarell

what is the actual application of fullerenes nowadays?

Damian

That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.

Tarell

Join the discussion...

what is the Synthesis, properties,and applications of carbon nano chemistry

Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.

In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google

Azam

anybody can imagine what will be happen after 100 years from now in nano tech world

Prasenjit

after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments

Azam

name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world

Prasenjit

how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?

Damian

silver nanoparticles could handle the job?

Damian

not now but maybe in future only AgNP maybe any other nanomaterials

Azam

Hello

Uday

I'm interested in Nanotube

Uday

this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15

Prasenjit

Join the discussion...

how did you get the value of 2000N.What calculations are needed to arrive at it